Is kinetic energy just as arbitrary as potential energy?

Potential energy $U$ can be defined up to an arbitrary additive constant $c$ because $$F=-\dfrac{d(U+c)}{dx}=-\dfrac{dU}{dx}=ma$$ And therefore the equation of motion remains unchanged. I think the same holds for kinetic energy $T$ using a similar reasoning and I want to make sure that I'm getting it right.

In a system where conservation of mechanical energy holds true $$T+U=\dfrac{1}{2}mv^2+U=\text{constant}$$ differentiating with respect to position $x$ we get $$\dfrac{dT}{dx}=-\dfrac{dU}{dx}=F=ma$$ therefore by the same token one expects that kinetic energy can be defined up to an arbitrary additive constant such that $$T_c=\dfrac{1}{2}mv^2+c$$ where usually we prefer to set $c=0$ for simplicity. Is this fact right about kinetic energy?

• $T=\frac{1}{2}m\left(\frac{dx}{dt}\right)^2$ does not depend on $x$, so in your computation $\frac{dT}{dx}$ should be 0! See @Jon for the right way to do it.
– user154997
Jun 13 '17 at 10:33
• @LucJ.Bourhis Nope it does not. $$\dfrac{dT}{dx}=\dfrac{d[\dfrac{1}{2}mv^2]}{dx}=mv \dfrac{dv}{dx}=mv \dfrac{dv}{dt} \dfrac{dt}{dx}=mva\dfrac{1}{v}=ma=F$$ just as stated compactly in my question. Jun 13 '17 at 10:43
• That computation works only in one dimension… And really, in mechanics position and velocity have to be considered as independent variable. I mean you can independently fix the initial position and the initial velocity e.g.
– user154997
Jun 13 '17 at 10:49
• Mathematics aside, isn't it the case that in the rest frame of a particle, the kinetic (resulting from motion) energy of the particle is zero? This isn't arbitrary is it? Jun 13 '17 at 12:25
• @OmarNagib One must be careful when applying the chain rule: you cannot just insert differentials left and right. To start with, one must define what variable is $v$ function of (say, $t$); when so, one must then make sure that such function is invertible for the position in any point of the domain (it usually isn't) and so is its inverse derivatives. Jun 13 '17 at 13:13

That's right but be careful about that constant that takes an interesting value from special relativity. Anyhow, I prefer this other approach. Consider $$m\frac{d{\bf v}}{dt}={\bf F}$$ and multiply both members of this equation by ${\bf v}$. You will get $$m{\bf v}\cdot\frac{d{\bf v}}{dt}={\bf F}\cdot{\bf v}$$ that means $$\frac{d}{dt}\left(\frac{1}{2}mv^2\right)={\bf F}\cdot{\bf v}.$$ You can integrate in time obtaining $$\frac{1}{2}mv^2+c=\int{\bf F}\cdot{\bf v}dt.$$ The constant is generally fixed by the problem at hand, e.g. by computing the work of the force on the right hand side.

• The work is by definition the integral of the right hand side along a path, it isn't the "primitive" of the differential form. As such, one obtains eventually that $W_{\gamma} = T(B) - T(A)$ and no such constant ever appears in the definition of kinetic energy, the latter being defined exactly as the quadratic form $T(x,y,z) = 1/2 m (\dot{x}^2+\dot{y}^2+\dot{z}^2)$. Jun 13 '17 at 10:44
• I have just integrated a differential equation aside from definitions.
– Jon
Jun 13 '17 at 11:30
• Yes, my objection is that the right hand side does not equal the work: it does so only if you integrated along a path. Jun 13 '17 at 12:54

My answer would be yes; in the same frame of reference the work done on the mass $m$ will be the same for every observer:

$$W_{1-2}=\int_{t_1}^{t_2}\vec{F}.\vec{v}dt= \int_{t_1}^{t_2}\frac{d}{dt}\left(\frac{mv^2}{2}\right)dt=\frac{mv_2^2}{2} - \frac{mv_1^2}{2} = \Delta E_{kin}$$

so therfore we can a time-independent constant to the function in the second integral - $\frac{mv^2}{2}$ - without changing the result. And because kinetic energy can be definied as this function, the kinetic energy is definied up to a constant.